Problem: Let $f(x) = 4x^{2}+6x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $4x^{2}+6x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 4, b = 6, c = -4$ $ x = \dfrac{-6 \pm \sqrt{6^{2} - 4 \cdot 4 \cdot -4}}{2 \cdot 4}$ $ x = \dfrac{-6 \pm \sqrt{100}}{8}$ $ x = \dfrac{-6 \pm 10}{8}$ $x =\frac{1}{2},-2$